Head Up Theory Question

[donkey69]

The unexpoitable strategy is not difficult to find. In fact its staring everyone in the face but they fail to realise it. Case of missing the forrest for the trees. The following strategy WILL DOMINATE ALL OTHER STRATEGIES in the long run. No ifs or buts. It will be EV+ against ANY other strategy. It will be EV=0 against itself.

Basically its a tight aggressive strategy:
If you draw 1: Fold
If you draw 2-999,999 check if you are first mover and if the other person bets call it.
If you draw 2-999,999 and are second mover, call the bet if there is one.
If you draw 1,000,000 raise infinitely until you are called.

Reasoning:
Folding the 1 is obvious since you are the dog with certainty.
Limiting the betting to maximum of 1 round for 2-999,9999: since you know you dont have the nuts but the other person may you try to limit your lossess as much as possible. On average you will beat your opponent half the time and he will beat you the other half. Should break even in the long run.
1,000,000 The Nuts: All bets made and reraises made are EV+ so you want as much action as possible.

I believe the above strategy will beat all the other strategies mentioned above. If both players are expert players betting should not proceed for more than 1 round.

[Stephen H]

[ QUOTE ]

Limiting the betting to maximum of 1 round for 2-999,9999: since you know you dont have the nuts but the other person may you try to limit your lossess as much as possible. On average you will beat your opponent half the time and he will beat you the other half. Should break even in the long run.


[/ QUOTE ]

This strategy is easily exploitable and is an obvious loser. The statement "Should break even in the long run" is incorrect; the times you're calling a bet you'll be behind a lot more often than 50/50; the times you're ahead, the action will go check/check and you'll win less than you lose in this range, on average. The bets you win on the nuts hand will not nearly make up for the huge number of bets you lose here, one bet at a time. Calling a bet with this many low cards is terrible.
Also realize that a strategy tailored to exploiting this one will ALWAYS fold to any bet or raise, as that means you have the nuts, so you will make almost nothing the one time in a million that you have the nuts.

[donkey69]

Ok. Assume that there are only 5 numbers, 1-5. Please illustrate how my strategy can be beaten. What decision rules would you use?

The fact that you dont make money is not of significance. In any game breaking even against expert players is in no way a bad result.

[Stephen H]

Sure, let's take your strategy at the 1-5 card game. Just to be clear, here's how I interpret your strategy:

Going first:
1: Check/fold
2-4: Check/call
5: Bet/raise forever.

Going second:
1: Check behind or fold to a bet.
2-4: Check behind or call a bet.
5: Bet or raise forever.

Here is my counter strategy that has +EV over yours:
Going first:
1-3: Check, fold to a bet.
4: Bet, fold to a raise.
5: Bet (raise forever, but that will never come up given your strategy)

Going second:
1-3: Check behind, fold to a bet.
4: Bet, fold to a bet.
5: Bet.

It should be fairly obvious that the person going second has a positional advantage, so I'll show the hand chart for your strategy going second. Suffice it to say that your EV is even worse going first.
The BB code eats extra whitespace, so I'm not sure the best way to display the table. The columns are:
My card / Your card / Result

1 / 2 / You win $1 (check/check)
1 / 3 / You win $1 (check/check)
1 / 4 / You win $1 (check/check)
1 / 5 / You win $1 (check/bet/fold)
2 / 1 / I win $1 (check/check)
2 / 3 / You win $1 (check/check)
2 / 4 / You win $1 (check/check)
2 / 5 / You win $1 (check/bet/fold)
3 / 1 / I win $1 (check/check)
3 / 2 / I win $1 (check/check)
3 / 4 / You win $1 (check/check)
3 / 5 / You win $1 (check/bet/fold)
4 / 1 / I win $1 (bet/fold)
4 / 2 / I win $2 (bet/call)
4 / 3 / I win $2 (bet/call)
4 / 5 / You win $2 (bet/raise/fold)
5 / 1 / I win $1 (bet/fold)
5 / 2 / I win $2 (bet/call)
5 / 3 / I win $2 (bet/call)
5 / 4 / I win $2 (bet/call)

Note that while you win 10 out of the 20 combinations, you only win more than the ante when you have the 5 and I have the 4. In the case where you're going first, you will never win a bet from me; I'll only bet when I know I'm going to win. If you add up the results, you'll see that across the 20 possible hands, you'll win $11 and lose $15 for an EV of -$4/20, or -0.20. The rate for you going first is -$5/20, or -0.25. Combining the two gives your strategy a total EV of -0.45/hand versus my counterstrategy. As you can see, your theory that you should "break even" on your hands 2-4 is false; while you win 6 of the 12 hands, you only win $6 on your wins, and lose $11, far from breaking even. And you don't make more than the ante when you hold the nuts, so you aren't covering up the difference there. So, the "correct" strategy isn't as simplistic as that.

[xpsyuvz]

edit: I was trying to save StephenH some time, but it looks like he already posted while I was writing this.

Donkey69,

I came up with a counter-strategy to yours to show how it can be exploited ( -- and my opinion is that any strategy can be exploited if the opponent knows the strategy that is used).

First, just to be clear: let’s say the game requires a $1 ante. And we’re using a deck of 5 cards (1-5) where 5 wins.

Strategy for player-A will be:
When in the first position and player-A has:
1 -- then he folds
2, 3, or 4 -- then he checks and then calls (if he is raised).
5 -- then he raises indefinitely.
When in the second potion:
Basically he does the same as above (where for 2, 3, or 4 he checks or calls if raised.)

Strategy for player-B will be:
When in the first position, then with card
1-- Fold
2-- Check. If raised then fold.
3-- Check. If raised then fold.
4-- Raise. If re-raised then fold.
5-- Raise indefinably.
When in the 2nd position, then:
Basically do the same as above except for
4 -- If raised then fold. If player-A checks, then raise.


To look at the expected value we can play out the 40 different scenarios (20 where player-A is in the first position and 20 where player-A is in the second position). Here I’ll list the card combinations:

A in the 1st position vs. B in the 2nd
1 vs. 2 A loses -1 & B wins +1
1 vs. 3 A loses -1 & B wins +1
1 vs. 4 A loses -1 & B wins +1
1 vs. 5 A loses -1 & B wins +1
2 vs. 1 A wins +1 & B loses -1
2 vs. 3 A loses -1 & B wins +1
2 vs. 4 A loses -2 & B wins +2
2 vs. 5 A loses -2 & B wins +2
3 vs. 1 A wins +1 & B loses -1
3 vs. 2 A wins +1 & B lose -1
3 vs. 4 A loses -2 & B wins +2
3 vs. 5 A loses -2 & B wins +2
4 vs. 1 A wins +1 & B loses -1
4 vs. 2 A wins +1 & B loses -1
4 vs. 3 A wins +1 & B loses -1
4 vs. 5 A loses -2 & B wins +2
5 vs. 1 A wins +1 & B loses -1
5 vs. 2 A wins + 1 & B loses -1
5 vs. 3 A wins +1 & B loses -1
5 vs. 4 A wins +1 & B loses -1
(So far player-A wins +10 and loses -15 while player-B wins +15 and loses -10. And note how player-A’s losses against player-B’s 4 card outweigh his 4 card when up against B other cards.)

A in the 2nd position and B in the 1st position:
(I’m not going to bother typing this out but I think the results are that player-A wins +11 and loses - 15. And the only difference here is player-B’s 4 card raises into player-A’s 5 before folding.)

So the total expected value outcome is player-A loses $9.00 while player-B wins $9.00. (Here player-A doesn't breakeven like you suggested.)

And meanwhile I'm sure player-B's strategy is exploitable too... (If someone has an unexploitably strategy here, I'd be interested in hearing about it.)

[Stephen H]

Alright, I've been thinking about this a while now, and reading the previous posts, as well as re-reading the fabulous [0,1] game posts by Chen/Ankenman. Almost all the math I've lifted directly from the [0,1] posts, so please correct me where I've missapplied it. Here's what I see, starting with analysis of the fixed limit game:

If the game allows no folding and no checkraises, this is exactly game #3, with the result that you should raise with the top sqrt(2)-1 percent hands (Labeled the "Golden Mean of Poker" in the [0,1] posts) that you put your opponent on. Working out a chart, I get that with the top 2 cards in the game, you should be willing to put in the 15th bet; for the 16th bet you need the top card. This means that, as stated, you would bet and then put in 7 raises on top of that for a total of 16 bets back to you after your opponent re-raises your last raise. At this point you would just call.

First question: Since the cards are discrete values, should I be rounding in this table at each step? Or, since a strategy could be, say, when comtemplating making it 10 bets, to raise it with the top 148 cards, and also raise the 149th 68% of the time, is rounding simply adding error? I suspect there's no need to round, and as was said before, he who has the most accurate calculator (and randomization device, I guess!) has the +EV.

Once you allow folding, you introduce the bluff. The bluff rate needs to be 1 bluff raise for every p value raises, where p is the size of the pot after you raise. However, the range of value raising hands does not change; you still raise those the same amount. So, for each bet, you can add in a # of bluffs equal to (# of raising hands)/(value of pot after your raise). As the # of hands is going down quickly, and the value of the pot is going up, this becomes less than 1 around the 14th bet.
2 interesting points I've found.
First, since your opponent may be bluffing, his hand range is a little larger than before..and this actually adds up to enough to allow you to put in the 16th bet with the top 2 cards, one more bet than before (although it doesn't change the answer to the question as stated, since it's an even bet #). I also calculate that you can value raise about 14% of the time for the 17th bet.
Second, while you stop value raising with the 999,999 after 17 bets, the bluff rates go on forever (but are remarkably small), so you might be bluffing with the 999,999. I'm not entirely sure how the bluffing works once you reach the degenerative case of only value betting the 1,000,000, but I'm fairly sure that you'll be bluffing a fairly low percentage of the time past this point.

Questions: How often SHOULD you be bluffing once you reach the degenerative case? And, again, should I be doing rounding here? When I calculate the range of hands you put your opponent on for use with the "Golden Mean of Poker", should I be including the amount of bluffs? For example:
When the first bet goes in, you should have at least the top 414,213 cards (I can see why the [0,1] game was written as lowball - so you can say "top X cards" and "X or better" without confusion). You should also be bluffing 25% of the time, so you bet out with effectively 552,284.75 hands out of 1,000,000. Do you re-raise for value with the top 171,572 hands (.414 times 414,213) or the top 228,763 hands (.414 times 552,284)? I've assumed for my numbers above that you do include the bluffing range.

For pot limit, the first thing to note is that pot size/bet size considerations only affect bluff/fold percentages, and not value raise percentages. If you're ahead, you're ahead, and your raise will only gain you 1:1 on the value of your raise no matter how much is in the pot. Now, the bluff rate should be constant, because you're always offering the same pot odds to call; 2:1. Because of this, the hand range stays large a little longer and takes longer to converge to the top 1 card. According to my calculations, you're now value betting 999,999 up to 26 bets, value betting the 27th bet 75% of the time, and value betting the 28th bet 8.8% of the time. After that, the only raises going in with 999,999 should be bluffs. The bluff factor seems more important here, but again, I'm not sure just how we're supposed to bluff once we reach the degenerative case.

I haven't bothered to figure out any of the fold/call percentages, mainly because they would be a lot more work (for me, anyways..maybe it's trivial!) and aren't really required for this part of the problem. I'd like a little confirmation that I'm on the right track before I bothered to work on those parts.

To construct the tables, I used the following formulas (rather than put the whole tables in here)
Fixed Limit:
Nth bet
Pot size after raise = Pot size of (n-1)th bet + 2
Raising range = (sqrt(2)-1)*(Opponent's Hand Range)
Bluffing range = (Raising Range) / (Pot size after raise)
Opponent's Hand Range = (n-1)th Raising range + (n-1)th Bluffing range

1st bet: pot size = 3, Opponent hand range = 1,000,000

Pot Limit:
Nth bet:
Raising range = (sqrt(2)-1)*(Opponent's Hand Range)
Bluffing range = (Raising Range) / 2
Opponent's Hand Range = (n-1)th Raising range + (n-1)th Bluffing range

1st bet: Opponent's Hand Range = 1,000,000

Comments/Criticisms?

[Underlord]

[ QUOTE ]
If the game allows no folding and no checkraises, this is exactly game #3, with the result that you should raise with the top sqrt(2)-1 percent hands (Labeled the "Golden Mean of Poker" in the [0,1] posts) that you put your opponent on. Working out a chart, I get that with the top 2 cards in the game, you should be willing to put in the 15th bet; for the 16th bet you need the top card.

[/ QUOTE ]

The $1 ante also deviates from the [0,1] game.

Assuming no ante, no folding, and no checkraises it's exactly the Bill Chen [0,1] 3rd game (part 3).

http://tinyurl.com/auj5

"What this means is that when infinite bets are allowed, you should generally put in another raise with about 41% of the hands your opponent can have to have raised you this far."

From the indifference equation:
x_n = r^(n-1)/(1+2r)
r = 0.414 "Golden Mean of Poker"
x_n = point where opponent is indifferent to checking or betting
n = number of bets to be put in before stopping

This gives the optimal betting strategy for any given hand in this game:
h = hand # (ie 999,999)
n = number of raises where u should stop
ln (1- (1.828 * h/1000000)) / ln(0.414) + 1 = n

Since this game only has 1,000,000 values, the chart is short:
(how many bets should be put in with top x hands)
1 = 547,000
3 = 93,000
5 = 16,000
7 = 2000
9 = 470
11 = 80
13 = 13
15 = 2
17 = 0.4

Only odd numbers because check-raises are not allowed and you act first.

You put in the 15th bet with your top 2 hands (999,998 and 999,999). A 16th bet requires you having the top hand of 1,000,000. This assumes your opponent is also playing using optimal strategy (expert).

Answer: Put in the 15th bet and then just call.

[Stephen H]

[ QUOTE ]
The $1 ante also deviates from the [0,1] game.

[/ QUOTE ]

I disagree. One justification for no folding was "infinite pot"; and the size of the pot never comes into play in any of the equilibrium equations. The ante matters not at all. The only difference I see is that the [0,1] game is played on the real line, and the 1,000,000 card game is discrete.

[ QUOTE ]

"What this means is that when infinite bets are allowed, you should generally put in another raise with about 41% of the hands your opponent can have to have raised you this far."

From the indifference equation:
x_n = r^(n-1)/(1+2r)
r = 0.414 "Golden Mean of Poker"
x_n = point where opponent is indifferent to checking or betting
n = number of bets to be put in before stopping


[/ QUOTE ]

Of course! I had it mostly right, except for the starting bet - it's special, since you don't have to call a bet, but could check and get checked behind or bet to (where you have to just call). This changes the range for the initial bet.
Incidentally, I wouldn't go through the trouble of inverting this equation (while rounding to 3 sig figs in the process) when what you really want is a chart of what x_n is for varying values of n. Also of note is that the equation for y_n where n is even (player y goes 2nd) is identical to x_n, so you can create the chart with both even and odd n, and use it for going first and second.

I created a chart based on those equations, and now I get the same answers you have (although mine aren't rounded to 3 sig figs). Which puts the answer at "1 bet and 7 raises" for the game with no folding.

I think the logic on my extrapolation into the game with folding is quite faulty, and I no longer have any confidence in it. However, I'm fairly sure it's within an order of magnitude or so.

Thanks for the reply!

[Jacob_Juul]

maybe i'm totally off on this one...
but just to compare to something
there are a total of 1.000.000 hands.
in a regular hold em game there are a total of 196 hands...
1.000.000/196 = about 5102
there fore hand number 999.999 is equal to AA
Since AA lies in the interval of 994.898 - 1.000.000 (1.000.000-5102=994.898)

[Slade]

On the issue of if you should ever fold 999,999, I think it is dependant on if you are raising 999,998 the same amount of times as 999,999. If you are then you should never fold 999,999. If not then some small amount dictated by the pot size should be correct. Note that this is for two reasons. One being that you would rather choose to fold something lower than 999,999 if you could have something smaller in the same situation. The second being if your last value raise with 999,999 is something you would only make with 999,999+ then your opponent couldnt be value raising with a lower range of numbers. (If an opponent is making a value raise with 999,998+ then you should always respond with a call when you have 999,999. However this situation would be impossible if your last value raise came with 999,999+)